Can you answer these queries III (line segment tree to find the maximum sub-segment sum of the interval)

Can you answer these queries III (the maximum sub-segment sum of the interval)

Blue book title Can you answer these queries III
Given a sequence A of length N, and M instructions (N≤500000, M≤100000), each instruction may be one of the following two:

• "2 xy", change A[x] to y.
• "1 xy", query the largest continuous sub-segment sum in the interval [x,y], that is, max(x≤l≤r≤y)∑ri=lA[i].

​ For each query, output an integer to indicate the answer.

Idea: In the
online segment tree, except for the interval endpoints, in the maintenance interval and $sum$ , the largest sub-segment of the interval and$dat$ , the largest continuous sub-segment at the left end and$lmax$ , the largest continuous sub-segment at the right end and$rmax$
transfer equation changes:
$sum\left(p\right)=sum(p\ast 2)+sum(p\ast 2+1)$
$rmax(p)=max(rmax(p\ast 2+1),sum(p\ast 2+1)+rmax(p\ast 2))$
$lmax(p)=max(lmax(p\ast 2),lmax(p\ast 2+1)+sum(2\ast p))$
$dat\left(p\right)=max\left(dat\right(p\ast 2),max(dat(p\ast 2+1),lmax(p\ast 2+1)+rmax(p\ast 2)))$

struct SegmentTree{
    
    
	int l,r;
	int dat,sum,lmax,rmax;
	#define l(x) tree[x].l
    #define r(x) tree[x].r
    #define dat(x) tree[x].dat
    #define sum(x) tree[x].sum
    #define lmax(x) tree[x].lmax
    #define rmax(x) tree[x].rmax
}tree[maxn*4];
int a[maxn];
int n,m;

void build(int p,int l,int r){
    
    
	l(p)=l,r(p)=r;
	if(l==r){
    
    
        dat(p)=a[l];sum(p)=a[l];lmax(p)=a[l];rmax(p)=a[l];return;
    }
	int mid=(l+r)/2;
	build(p*2,l,mid);
	build(p*2+1,mid+1,r);
	sum(p)=sum(p*2)+sum(p*2+1);
	rmax(p)=max(rmax(p*2+1),sum(p*2+1)+rmax(p*2));
	lmax(p)=max(lmax(p*2),lmax(p*2+1)+sum(2*p));
	dat(p)=max(dat(p*2),max(dat(p*2+1),lmax(p*2+1)+rmax(p*2)));
}

void change(int p,int x,int z){
    
    
	if(l(p)==r(p)){
    
    
		dat(p)=z;sum(p)=z;lmax(p)=z;rmax(p)=z;return;
	}
	int mid=(l(p)+r(p))/2;
	if(x<=mid) change(p*2,x,z);
    else change(p*2+1,x,z);
	sum(p)=sum(p*2)+sum(p*2+1);
	rmax(p)=max(rmax(p*2+1),sum(p*2+1)+rmax(p*2));
	lmax(p)=max(lmax(p*2),lmax(p*2+1)+sum(2*p));
	dat(p)=max(dat(p*2),max(dat(p*2+1),lmax(p*2+1)+rmax(p*2)));
}

SegmentTree ask(int p,int l,int r){
    
    
	if(l<=l(p)&&r>=r(p)) return tree[p];
	SegmentTree a,b,ans;
	a.sum=a.lmax=a.rmax=a.dat=b.sum=b.lmax=b.rmax=b.dat=-1*inf;
	ans.sum=0;
    int mid=(l(p)+r(p))/2;
    if(l<=mid){
    
    a=ask(p*2,l,r);ans.sum+=a.sum;}
    if(r>mid){
    
    b=ask(p*2+1,l,r);ans.sum+=b.sum;}
    ans.dat=max(max(a.dat,b.dat),a.rmax+b.lmax);
    ans.lmax=max(a.lmax,a.sum+b.lmax);
    ans.rmax=max(b.rmax,b.sum+a.rmax);
    if(l>mid)ans.lmax=max(ans.lmax,b.lmax);
    if(r<=mid)ans.rmax=max(ans.rmax,a.rmax);
    return ans;
}

int main(){
    
    
	cin>>n>>m;
	for(int i=1;i<=n;i++)
		cin>>a[i];
	build(1,1,n);
	while(m--){
    
    
        int op,x,y;
        cin>>op>>x>>y;
		if(op==2){
    
    
			change(1,x,y);
		}
		else {
    
    if(x>y)swap(x,y);cout<<ask(1,x,y).dat<<endl;}
	}
}